\(A=\dfrac{x^2+2x-3}{x+3}=\dfrac{x^2+3x-x-3}{x+3}=\dfrac{\left(x+3\right)\left(x-1\right)}{x+3}=x-1\)

\(A=\dfrac{x^2+2x-3}{x+3}=\dfrac{x^2-x+3x-3}{x+3}=\dfrac{x\left(x-1\right)+3\left(x-1\right)}{x+3}=\dfrac{\left(x-1\right)\left(x+3\right)}{x+3}=x-1\)

\(\dfrac{x^3+64}{2x^3-8x^2+32x}\\ =\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\\ =\dfrac{x+4}{2x}\)

\(\dfrac{x^3+64}{2x^3-8x^3+32x}\)

\(=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\)

\(=\dfrac{x+4}{2x}\)

\(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\dfrac{\left(x+y\right)\left(x+2y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}\)

\(=\dfrac{x+y}{x^2-y^2}\)

\(=\dfrac{1}{x-y}\)

\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)

b: \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)

c: \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\)

a: \(A=\dfrac{3\left(1-2x\right)}{2x\left(x^2+1\right)-\left(x^2+1\right)}\)

\(=\dfrac{-3\left(2x-1\right)}{\left(x^2+1\right)\left(2x-1\right)}=\dfrac{-3}{x^2+1}\)

b: Khi x=3 thì \(A=\dfrac{-3}{3^2+1}=-\dfrac{3}{10}\)

c: x^2+1>=0

=>3/x^2+1>=0

=>-3/x^2+1<=0

=>A<=0(ĐPCM)

\(\frac{2\left(x-2\right)\left(x-3\right)}{\left(3x^2-27\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{\left(x-2\right)3\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)

\(=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+x^2\right)+\left(2x^2+2x+2\right)}=\frac{x^3.\left(x-1\right)-\left(x-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)

\(=\frac{\left(x^3-1\right).\left(x-1\right)}{\left(x^2+2\right)\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2.\left(x^2+x+1\right)}{\left(x^2+2\right)\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2}{x^2+2}\)

a kham khảo nha , e nhờ a e lm chứ ko phải e lm nha ! 

\(\left(x-2\right)\left(\frac{3}{x}+2-\frac{5}{2x}-4+\frac{8}{x^2}-4\right)\)

\(\left(x-2\right)\left[\left(\frac{3}{x}-\frac{5}{2x}\right)-6+\frac{8}{x^2}\right]\)

\(\left(x-2\right)\left(\frac{1}{2x}-6+\frac{8}{x^2}\right)\)

\(\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)\)

\(=\left(x-2\right)\left[\frac{3}{x+2}-\frac{5}{2\left(x-2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{3.2\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{8.2}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)-5\left(x+2\right)+16}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\frac{\left(x-2\right)\left(x-6\right)}{2\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-6}{2\left(x+2\right)}\)

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)